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Expressing sin^6 q in the form a + bcos 2q + ccos 4q + dcos 6q using De Moivre's watch

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    where a, b, c, d are constants to be found.

    I have no idea how to go about this?..
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    PS Reviewer
    use the z+1/z and z-1/z relation
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    (Original post by cpdavis)
    use the z+1/z and z-1/z relation
    Sorry but it's not the book, could you elaborate?
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    Start with:

    2^6sin^6x = (z - \frac{1}{z})^6

    Can you see what to do from there?
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    PS Reviewer
    ok, sorry for being brief

    from de movires therom, we are able to state that z-1/z=sin (theta)

    there is a really clear explanation about this in the AQA online textbook, here is the link:

    http://store.aqa.org.uk/qual/pdf/AQA-MFP2-TEXTBOOK.PDF

    P63

    hope it helps
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    (Original post by soutioirsim)
    Start with:

    sin^6x = (z - \frac{1}{z})^6

    Can you see what to do from there?
    ..
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    (Original post by cpdavis)
    ok, sorry for being brief

    from de movires therom, we are able to state that z-1/z=sin (theta)

    there is a really clear explanation about this in the AQA online textbook, here is the link:

    http://store.aqa.org.uk/qual/pdf/AQA-MFP2-TEXTBOOK.PDF

    P63

    hope it helps
    Thank you
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    (z-\dfrac{1}{z})^6=(2i \sin q)^6

    Remember, it's 2i \sin q and not just \sin q - forgetting this could have apocalyptic consequences.

    Edit: Where z=\cos q + i\sin q...
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    Someone should also really mention what you're defining z to be, too. If the OPs never seen this before then making him guess like that is a little cruel. :p:
 
 
 

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